Finite Mathematics Quiz: 2x2 Matrix Inverses
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2x2 Matrix InversesQuestion 1 of 14

The inverse of the matrix A=(3x46)A = \begin{pmatrix} 3 & x \\ 4 & 6 \end{pmatrix} is A−1=(−32.52−1.5)A^{-1} = \begin{pmatrix} -3 & 2.5 \\ 2 & -1.5 \end{pmatrix}. What is the value of xx?

55
−5-5
44
2.52.5
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Finite Mathematics Quiz

Finite Mathematics Quiz: 2x2 Matrix Inverses

Practice 2x2 Matrix Inverses in Finite Mathematics with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

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This quiz focuses on 2x2 Matrix Inverses, giving you a quick way to practice the rules, question types, and explanations that matter most for Finite Mathematics.

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Question 1

The inverse of the matrix A=(3x46)A = \begin{pmatrix} 3 & x \\ 4 & 6 \end{pmatrix} is A−1=(−32.52−1.5)A^{-1} = \begin{pmatrix} -3 & 2.5 \\ 2 & -1.5 \end{pmatrix}. What is the value of xx?

  1. 55 (correct answer)
  2. −5-5
  3. 44
  4. 2.52.5
Explanation: The product of a matrix and its inverse is the identity matrix, i.e., AA−1=I=(1001)AA^{-1} = I = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}. We can use this property to solve for xx. Let's compute the entry in the first row, second column of the product AA−1AA^{-1}: (3x46)(−32.52−1.5)\begin{pmatrix} 3 & x \\ 4 & 6 \end{pmatrix} \begin{pmatrix} -3 & 2.5 \\ 2 & -1.5 \end{pmatrix}. The entry is (3)(2.5)+(x)(−1.5)=7.5−1.5x(3)(2.5) + (x)(-1.5) = 7.5 - 1.5x. This must be equal to the corresponding entry in the identity matrix, which is 0. So, we set up the equation 7.5−1.5x=07.5 - 1.5x = 0. Solving for xx gives 1.5x=7.51.5x = 7.5, which means x=5x = 5.

Question 2

For what value of cc does the linear system represented by the matrix equation (2−6−3c)(xy)=(51)\begin{pmatrix} 2 & -6 \\ -3 & c \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} 5 \\ 1 \end{pmatrix} fail to have a unique solution?

  1. −9-9
  2. 99 (correct answer)
  3. −6-6
  4. 33
Explanation: A system of linear equations Ax=bA\mathbf{x}=\mathbf{b} has a unique solution if and only if the coefficient matrix AA is invertible. The system fails to have a unique solution if AA is singular, which means its determinant is zero. The coefficient matrix is A=(2−6−3c)A = \begin{pmatrix} 2 & -6 \\ -3 & c \end{pmatrix}. We calculate its determinant: det⁡(A)=(2)(c)−(−6)(−3)=2c−18\det(A) = (2)(c) - (-6)(-3) = 2c - 18. To find the value of cc for which there is no unique solution, we set the determinant to zero: 2c−18=02c - 18 = 0. Solving for cc gives 2c=182c = 18, so c=9c = 9.

Question 3

Let A=(1−2−37)A = \begin{pmatrix} 1 & -2 \\ -3 & 7 \end{pmatrix}. If its inverse is A−1=(abcd)A^{-1} = \begin{pmatrix} a & b \\ c & d \end{pmatrix}, what is the value of the expression a−da - d?

  1. −6-6
  2. 88
  3. 6/136/13
  4. 66 (correct answer)
Explanation: First, compute the inverse of matrix AA. The determinant is det⁡(A)=(1)(7)−(−2)(−3)=7−6=1\det(A) = (1)(7) - (-2)(-3) = 7 - 6 = 1. The formula for the inverse of a 2×22 \times 2 matrix (wxyz)\begin{pmatrix} w & x \\ y & z \end{pmatrix} is 1wz−xy(z−x−yw)\frac{1}{wz-xy} \begin{pmatrix} z & -x \\ -y & w \end{pmatrix}. Applying this to AA, we get A−1=11(7−(−2)−(−3)1)=(7231)A^{-1} = \frac{1}{1} \begin{pmatrix} 7 & -(-2) \\ -(-3) & 1 \end{pmatrix} = \begin{pmatrix} 7 & 2 \\ 3 & 1 \end{pmatrix}. By comparing this to (abcd)\begin{pmatrix} a & b \\ c & d \end{pmatrix}, we identify a=7a=7, b=2b=2, c=3c=3, and d=1d=1. The required expression is a−d=7−1=6a - d = 7 - 1 = 6.

Question 4

The matrix equation (3k12)(xy)=(53)\begin{pmatrix} 3 & k \\ 1 & 2 \end{pmatrix}\begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} 5 \\ 3 \end{pmatrix} has solution x=1x = 1. For what value of kk does y=1y = 1 as well?

  1. k=2k = 2 (correct answer)
  2. k=1k = 1
  3. k=3k = 3
  4. k=4k = 4
Explanation: If x=1x = 1 and y=1y = 1, then substituting into the matrix equation: (3k12)(11)=(53)\begin{pmatrix} 3 & k \\ 1 & 2 \end{pmatrix}\begin{pmatrix} 1 \\ 1 \end{pmatrix} = \begin{pmatrix} 5 \\ 3 \end{pmatrix}. This gives us (3+k1+2)=(53)\begin{pmatrix} 3 + k \\ 1 + 2 \end{pmatrix} = \begin{pmatrix} 5 \\ 3 \end{pmatrix}. From the first component: 3+k=53 + k = 5, so k=2k = 2. We can verify with the second component: 1+2=31 + 2 = 3 ✓. The other choices would give different values for the first component that don't equal 5.

Question 5

The system of linear equations 3x+5y=−13x + 5y = -1 and 2x+4y=22x + 4y = 2 is represented by the matrix equation Ax=bA\mathbf{x} = \mathbf{b}. The solution is given by x=A−1b\mathbf{x} = A^{-1}\mathbf{b}. What is the entry in the first row of the solution vector x\mathbf{x}?

  1. −7-7 (correct answer)
  2. 44
  3. −14-14
  4. 88
Explanation: First, identify the coefficient matrix A=(3524)A = \begin{pmatrix} 3 & 5 \\ 2 & 4 \end{pmatrix} and the constant vector b=(−12)\mathbf{b} = \begin{pmatrix} -1 \\ 2 \end{pmatrix}. To find the solution x=A−1b\mathbf{x} = A^{-1}\mathbf{b}, we must first compute A−1A^{-1}. The determinant is det⁡(A)=(3)(4)−(5)(2)=12−10=2\det(A) = (3)(4) - (5)(2) = 12 - 10 = 2. The inverse is A−1=12(4−5−23)A^{-1} = \frac{1}{2} \begin{pmatrix} 4 & -5 \\ -2 & 3 \end{pmatrix}. The solution is x=12(4−5−23)(−12)=12(4(−1)+(−5)(2)−2(−1)+3(2))=12(−148)=(−74)\mathbf{x} = \frac{1}{2} \begin{pmatrix} 4 & -5 \\ -2 & 3 \end{pmatrix} \begin{pmatrix} -1 \\ 2 \end{pmatrix} = \frac{1}{2} \begin{pmatrix} 4(-1) + (-5)(2) \\ -2(-1) + 3(2) \end{pmatrix} = \frac{1}{2} \begin{pmatrix} -14 \\ 8 \end{pmatrix} = \begin{pmatrix} -7 \\ 4 \end{pmatrix}. The entry in the first row of the solution vector is x=−7x = -7.

Question 6

Let matrices A=(2153)A = \begin{pmatrix} 2 & 1 \\ 5 & 3 \end{pmatrix} and C=(10−12)C = \begin{pmatrix} 1 & 0 \\ -1 & 2 \end{pmatrix}. If the matrix equation AX=CAX = C holds, what is the sum of the elements in the first column of matrix XX?

  1. −10-10
  2. 22
  3. −3-3 (correct answer)
  4. 55
Explanation: To solve for matrix XX in the equation AX=CAX = C, we multiply both sides on the left by A−1A^{-1}, yielding X=A−1CX = A^{-1}C. First, we must find the inverse of AA. The determinant of AA is det⁡(A)=(2)(3)−(1)(5)=6−5=1\det(A) = (2)(3) - (1)(5) = 6 - 5 = 1. The inverse is A−1=11(3−1−52)=(3−1−52)A^{-1} = \frac{1}{1} \begin{pmatrix} 3 & -1 \\ -5 & 2 \end{pmatrix} = \begin{pmatrix} 3 & -1 \\ -5 & 2 \end{pmatrix}. Next, we compute the product X=A−1C=(3−1−52)(10−12)=(3(1)+(−1)(−1)3(0)+(−1)(2)−5(1)+2(−1)−5(0)+2(2))=(4−2−74)X = A^{-1}C = \begin{pmatrix} 3 & -1 \\ -5 & 2 \end{pmatrix} \begin{pmatrix} 1 & 0 \\ -1 & 2 \end{pmatrix} = \begin{pmatrix} 3(1)+(-1)(-1) & 3(0)+(-1)(2) \\ -5(1)+2(-1) & -5(0)+2(2) \end{pmatrix} = \begin{pmatrix} 4 & -2 \\ -7 & 4 \end{pmatrix}. The elements in the first column of XX are 4 and -7. Their sum is 4+(−7)=−34 + (-7) = -3.

Question 7

Given the matrix equation XA=BXA = B, where A=(3−2−43)A = \begin{pmatrix} 3 & -2 \\ -4 & 3 \end{pmatrix} and B=(1123)B = \begin{pmatrix} 1 & 1 \\ 2 & 3 \end{pmatrix}, the solution for XX is found by computing X=BA−1X = BA^{-1}. Which of the following matrices is XX?

  1. (751813)\begin{pmatrix} 7 & 5 \\ 18 & 13 \end{pmatrix} (correct answer)
  2. (791013)\begin{pmatrix} 7 & 9 \\ 10 & 13 \end{pmatrix}
  3. (−11−65)\begin{pmatrix} -1 & 1 \\ -6 & 5 \end{pmatrix}
  4. (−1−325)\begin{pmatrix} -1 & -3 \\ 2 & 5 \end{pmatrix}
Explanation: To solve for XX, we must first calculate A−1A^{-1}. The determinant is det⁡(A)=(3)(3)−(−2)(−4)=9−8=1\det(A) = (3)(3) - (-2)(-4) = 9 - 8 = 1. The inverse is A−1=11(3243)=(3243)A^{-1} = \frac{1}{1} \begin{pmatrix} 3 & 2 \\ 4 & 3 \end{pmatrix} = \begin{pmatrix} 3 & 2 \\ 4 & 3 \end{pmatrix}. Now, we compute the product X=BA−1X = BA^{-1}: X=(1123)(3243)=(1(3)+1(4)1(2)+1(3)2(3)+3(4)2(2)+3(3))=(3+42+36+124+9)=(751813)X = \begin{pmatrix} 1 & 1 \\ 2 & 3 \end{pmatrix} \begin{pmatrix} 3 & 2 \\ 4 & 3 \end{pmatrix} = \begin{pmatrix} 1(3)+1(4) & 1(2)+1(3) \\ 2(3)+3(4) & 2(2)+3(3) \end{pmatrix} = \begin{pmatrix} 3+4 & 2+3 \\ 6+12 & 4+9 \end{pmatrix} = \begin{pmatrix} 7 & 5 \\ 18 & 13 \end{pmatrix}.

Question 8

Let AA be a 2×22 \times 2 matrix such that A−1=(−1/21/23/2−1/2)A^{-1} = \begin{pmatrix} -1/2 & 1/2 \\ 3/2 & -1/2 \end{pmatrix}. If the system of equations Ax=(46)A\mathbf{x} = \begin{pmatrix} 4 \\ 6 \end{pmatrix} is solved for x=(xy)\mathbf{x} = \begin{pmatrix} x \\ y \end{pmatrix}, what is the value of x+yx+y?

  1. 11
  2. 22
  3. 44 (correct answer)
  4. 66
Explanation: The solution to the system Ax=bA\mathbf{x} = \mathbf{b} is given by x=A−1b\mathbf{x} = A^{-1}\mathbf{b}. We compute: x=(−1/21/23/2−1/2)(46)=((−1/2)(4)+(1/2)(6)(3/2)(4)+(−1/2)(6))=(−2+36−3)=(13)\mathbf{x} = \begin{pmatrix} -1/2 & 1/2 \\ 3/2 & -1/2 \end{pmatrix} \begin{pmatrix} 4 \\ 6 \end{pmatrix} = \begin{pmatrix} (-1/2)(4) + (1/2)(6) \\ (3/2)(4) + (-1/2)(6) \end{pmatrix} = \begin{pmatrix} -2 + 3 \\ 6 - 3 \end{pmatrix} = \begin{pmatrix} 1 \\ 3 \end{pmatrix}. Therefore, x=1x = 1 and y=3y = 3, so x+y=1+3=4x + y = 1 + 3 = 4.

Question 9

A system of linear equations is given by 5x−3y=155x - 3y = 15 and 3x−2y=83x - 2y = 8. The solution to the system can be found using the matrix equation Ax=bA\mathbf{x}=\mathbf{b}. What is the value of yy in the solution?

  1. 66
  2. 55 (correct answer)
  3. −85-85
  4. 8585
Explanation: The system can be written as Ax=bA\mathbf{x}=\mathbf{b} with A=(5−33−2)A = \begin{pmatrix} 5 & -3 \\ 3 & -2 \end{pmatrix}, x=(xy)\mathbf{x} = \begin{pmatrix} x \\ y \end{pmatrix}, and b=(158)\mathbf{b} = \begin{pmatrix} 15 \\ 8 \end{pmatrix}. The solution is x=A−1b\mathbf{x} = A^{-1}\mathbf{b}. First, find A−1A^{-1}. The determinant is det⁡(A)=(5)(−2)−(−3)(3)=−10−(−9)=−1\det(A) = (5)(-2) - (-3)(3) = -10 - (-9) = -1. The inverse is A−1=1−1(−23−35)=(2−33−5)A^{-1} = \frac{1}{-1} \begin{pmatrix} -2 & 3 \\ -3 & 5 \end{pmatrix} = \begin{pmatrix} 2 & -3 \\ 3 & -5 \end{pmatrix}. Now, calculate the solution vector: x=(xy)=(2−33−5)(158)=(2(15)+(−3)(8)3(15)+(−5)(8))=(30−2445−40)=(65)\mathbf{x} = \begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} 2 & -3 \\ 3 & -5 \end{pmatrix} \begin{pmatrix} 15 \\ 8 \end{pmatrix} = \begin{pmatrix} 2(15) + (-3)(8) \\ 3(15) + (-5)(8) \end{pmatrix} = \begin{pmatrix} 30 - 24 \\ 45 - 40 \end{pmatrix} = \begin{pmatrix} 6 \\ 5 \end{pmatrix}. The value of yy is the second entry in the solution vector, which is 5.

Question 10

Matrix A=(2314)A = \begin{pmatrix} 2 & 3 \\ 1 & 4 \end{pmatrix} represents the coefficients of a system of equations. If the system has solution (xy)=(5−1)\begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} 5 \\ -1 \end{pmatrix}, what is the constant vector b\mathbf{b} in the system Ax=bA\mathbf{x} = \mathbf{b}?

  1. (71)\begin{pmatrix} 7 \\ 1 \end{pmatrix} (correct answer)
  2. (13−3)\begin{pmatrix} 13 \\ -3 \end{pmatrix}
  3. (104)\begin{pmatrix} 10 \\ 4 \end{pmatrix}
  4. (7−1)\begin{pmatrix} 7 \\ -1 \end{pmatrix}
Explanation: To find b\mathbf{b}, we multiply AA by the solution vector: $$A\mathbf{x} = \begin{pmatrix} 2 & 3 \ 1 & 4 \end{pmatrix} \begin{pmatrix} 5 \ -1 \end{pmatrix} = \begin{pmatrix} 2(5) + 3(-1) \ 1(5) + 4(-1) \end{pmatrix} = \begin{pmatrix} 10 - 3 \ 5 - 4 \end{pmatrix} = \begin{pmatrix} 7 \ 1 \end{pmatrix}

Question 11

Let P=(1201)P = \begin{pmatrix} 1 & 2 \\ 0 & 1 \end{pmatrix} and Q=(3011)Q = \begin{pmatrix} 3 & 0 \\ 1 & 1 \end{pmatrix}. What is the entry in the first row, second column of the matrix (PQ)−1(PQ)^{-1}?

  1. 2/32/3
  2. −2/3-2/3 (correct answer)
  3. −2-2
  4. 22
Explanation: To find (PQ)−1(PQ)^{-1}, we first compute the product PQPQ. PQ=(1201)(3011)=(1(3)+2(1)1(0)+2(1)0(3)+1(1)0(0)+1(1))=(5211)PQ = \begin{pmatrix} 1 & 2 \\ 0 & 1 \end{pmatrix} \begin{pmatrix} 3 & 0 \\ 1 & 1 \end{pmatrix} = \begin{pmatrix} 1(3)+2(1) & 1(0)+2(1) \\ 0(3)+1(1) & 0(0)+1(1) \end{pmatrix} = \begin{pmatrix} 5 & 2 \\ 1 & 1 \end{pmatrix}. Next, we find the inverse of this resulting matrix. The determinant is det⁡(PQ)=(5)(1)−(2)(1)=3\det(PQ) = (5)(1) - (2)(1) = 3. The inverse is (PQ)−1=13(1−2−15)=(1/3−2/3−1/35/3)(PQ)^{-1} = \frac{1}{3} \begin{pmatrix} 1 & -2 \\ -1 & 5 \end{pmatrix} = \begin{pmatrix} 1/3 & -2/3 \\ -1/3 & 5/3 \end{pmatrix}. The entry in the first row, second column is −2/3-2/3.

Question 12

If P=(2132)P = \begin{pmatrix} 2 & 1 \\ 3 & 2 \end{pmatrix} and Q=(5412)Q = \begin{pmatrix} 5 & 4 \\ 1 & 2 \end{pmatrix} , what is the determinant of (PQ)−1(PQ)^{-1}?

  1. 16\frac{1}{6} (correct answer)
  2. 66
  3. 136\frac{1}{36}
  4. 3636
Explanation: Using the property that det⁡((PQ)−1)=1det⁡(PQ)=1det⁡(P)det⁡(Q)\det((PQ)^{-1}) = \frac{1}{\det(PQ)} = \frac{1}{\det(P)\det(Q)}. We have det⁡(P)=2(2)−1(3)=4−3=1\det(P) = 2(2) - 1(3) = 4 - 3 = 1 and det⁡(Q)=5(2)−4(1)=10−4=6\det(Q) = 5(2) - 4(1) = 10 - 4 = 6. Therefore det⁡(PQ)=det⁡(P)det⁡(Q)=1⋅6=6\det(PQ) = \det(P)\det(Q) = 1 \cdot 6 = 6, so det⁡((PQ)−1)=16\det((PQ)^{-1}) = \frac{1}{6}. Choice B gives det⁡(PQ)\det(PQ) instead of its reciprocal. Choice C incorrectly squares the reciprocal. Choice D incorrectly squares the determinant.

Question 13

Let A=(k5−1−k)A = \begin{pmatrix} k & 5 \\ -1 & -k \end{pmatrix}. If the inverse of matrix AA is equal to matrix AA itself (i.e., A−1=AA^{-1} = A), what is the value of k2k^2?

  1. 44
  2. 55
  3. 66 (correct answer)
  4. 2626
Explanation: If A−1=AA^{-1} = A, then multiplying by AA on the left gives AA−1=A2A A^{-1} = A^2, which means I=A2I = A^2, where II is the identity matrix. Let's compute A2A^2: A2=(k5−1−k)(k5−1−k)=(k(k)+5(−1)k(5)+5(−k)−1(k)+(−k)(−1)−1(5)+(−k)(−k))=(k2−500k2−5)A^2 = \begin{pmatrix} k & 5 \\ -1 & -k \end{pmatrix} \begin{pmatrix} k & 5 \\ -1 & -k \end{pmatrix} = \begin{pmatrix} k(k) + 5(-1) & k(5) + 5(-k) \\ -1(k) + (-k)(-1) & -1(5) + (-k)(-k) \end{pmatrix} = \begin{pmatrix} k^2 - 5 & 0 \\ 0 & k^2 - 5 \end{pmatrix}. We set this equal to the identity matrix I=(1001)I = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}. Equating the diagonal entries gives the equation k2−5=1k^2 - 5 = 1. Solving for k2k^2 yields k2=6k^2 = 6.

Question 14

If A=(3142)A = \begin{pmatrix} 3 & 1 \\ 4 & 2 \end{pmatrix}, what is the sum of all elements of the matrix A+A−1A + A^{-1}?

  1. 5.55.5
  2. 10.510.5 (correct answer)
  3. 11.511.5
  4. 12.512.5
Explanation: First, we need to find the inverse of matrix AA. The determinant of AA is det⁡(A)=(3)(2)−(1)(4)=6−4=2\det(A) = (3)(2) - (1)(4) = 6 - 4 = 2. The inverse is A−1=12(2−1−43)=(1−0.5−21.5)A^{-1} = \frac{1}{2} \begin{pmatrix} 2 & -1 \\ -4 & 3 \end{pmatrix} = \begin{pmatrix} 1 & -0.5 \\ -2 & 1.5 \end{pmatrix}. Next, we compute the sum of the matrices A+A−1A + A^{-1}: A+A−1=(3142)+(1−0.5−21.5)=(3+11−0.54−22+1.5)=(40.523.5)A + A^{-1} = \begin{pmatrix} 3 & 1 \\ 4 & 2 \end{pmatrix} + \begin{pmatrix} 1 & -0.5 \\ -2 & 1.5 \end{pmatrix} = \begin{pmatrix} 3+1 & 1-0.5 \\ 4-2 & 2+1.5 \end{pmatrix} = \begin{pmatrix} 4 & 0.5 \\ 2 & 3.5 \end{pmatrix}. Finally, we find the sum of all elements of this resulting matrix: 4+0.5+2+3.5=10.54 + 0.5 + 2 + 3.5 = 10.5.